Minimum Window Substring LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Minimum Window Substring in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Minimum Window Substring– LeetCode Problem

Minimum Window Substring– LeetCode Problem

Problem:

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

• m == s.length
• n == t.length
• 1 <= m, n <= 105
• s and t consist of uppercase and lowercase English letters.

Minimum Window Substring– LeetCode Solutions

Minimum Window Substring in C++:

class Solution {
 public:
  string minWindow(string s, string t) {
    vector<int> count(128);
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (const char c : t)
      ++count[c];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s[r]] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 < minLength) {
          bestLeft = l;
          minLength = r - l + 1;
        }
        if (++count[s[l++]] > 0)
          ++required;
      }
    }

    return bestLeft == -1 ? "" : s.substr(bestLeft, minLength);
  }
};

Minimum Window Substring in Java:

class Solution {
  public String minWindow(String s, String t) {
    int[] count = new int[128];
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (final char c : t.toCharArray())
      ++count[c];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 < minLength) {
          bestLeft = l;
          minLength = r - l + 1;
        }
        if (++count[s.charAt(l++)] > 0)
          ++required;
      }
    }

    return bestLeft == -1 ? "" : s.substring(bestLeft, bestLeft + minLength);
  }
}

Minimum Window Substring in Python:

class Solution:
  def minWindow(self, s: str, t: str) -> str:
    count = Counter(t)
    required = len(t)
    bestLeft = -1
    minLength = len(s) + 1

    l = 0
    for r, c in enumerate(s):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      while required == 0:
        if r - l + 1 < minLength:
          bestLeft = l
          minLength = r - l + 1
        count[s[l]] += 1
        if count[s[l]] > 0:
          required += 1
        l += 1

    return '' if bestLeft == -1 else s[bestLeft: bestLeft + minLength]